3.201 \(\int \frac{(e+f x) \sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx\)

Optimal. Leaf size=175 \[ \frac{i f \sinh ^2(c+d x)}{4 a d^2}-\frac{f \sinh (c+d x)}{a d^2}+\frac{2 i f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}+\frac{(e+f x) \cosh (c+d x)}{a d}-\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x) \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{3 i e x}{2 a}+\frac{3 i f x^2}{4 a} \]

[Out]

(((3*I)/2)*e*x)/a + (((3*I)/4)*f*x^2)/a + ((e + f*x)*Cosh[c + d*x])/(a*d) + ((2*I)*f*Log[Cosh[c/2 + (I/4)*Pi +
(d*x)/2]])/(a*d^2) - (f*Sinh[c + d*x])/(a*d^2) - ((I/2)*(e + f*x)*Cosh[c + d*x]*Sinh[c + d*x])/(a*d) + ((I/4)
*f*Sinh[c + d*x]^2)/(a*d^2) - (I*(e + f*x)*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

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Rubi [A]  time = 0.262787, antiderivative size = 175, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.241, Rules used = {5557, 3310, 3296, 2637, 3318, 4184, 3475} \[ \frac{i f \sinh ^2(c+d x)}{4 a d^2}-\frac{f \sinh (c+d x)}{a d^2}+\frac{2 i f \log \left (\cosh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )\right )}{a d^2}+\frac{(e+f x) \cosh (c+d x)}{a d}-\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{d x}{2}+\frac{i \pi }{4}\right )}{a d}-\frac{i (e+f x) \sinh (c+d x) \cosh (c+d x)}{2 a d}+\frac{3 i e x}{2 a}+\frac{3 i f x^2}{4 a} \]

Antiderivative was successfully verified.

[In]

Int[((e + f*x)*Sinh[c + d*x]^3)/(a + I*a*Sinh[c + d*x]),x]

[Out]

(((3*I)/2)*e*x)/a + (((3*I)/4)*f*x^2)/a + ((e + f*x)*Cosh[c + d*x])/(a*d) + ((2*I)*f*Log[Cosh[c/2 + (I/4)*Pi +
(d*x)/2]])/(a*d^2) - (f*Sinh[c + d*x])/(a*d^2) - ((I/2)*(e + f*x)*Cosh[c + d*x]*Sinh[c + d*x])/(a*d) + ((I/4)
*f*Sinh[c + d*x]^2)/(a*d^2) - (I*(e + f*x)*Tanh[c/2 + (I/4)*Pi + (d*x)/2])/(a*d)

Rule 5557

Int[(((e_.) + (f_.)*(x_))^(m_.)*Sinh[(c_.) + (d_.)*(x_)]^(n_.))/((a_) + (b_.)*Sinh[(c_.) + (d_.)*(x_)]), x_Sym
bol] :> Dist[1/b, Int[(e + f*x)^m*Sinh[c + d*x]^(n - 1), x], x] - Dist[a/b, Int[((e + f*x)^m*Sinh[c + d*x]^(n
- 1))/(a + b*Sinh[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3318

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[(2*a)^n, Int[(c
+ d*x)^m*Sin[(1*(e + (Pi*a)/(2*b)))/2 + (f*x)/2]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
- b^2, 0] && IntegerQ[n] && (GtQ[n, 0] || IGtQ[m, 0])

Rule 4184

Int[csc[(e_.) + (f_.)*(x_)]^2*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Simp[((c + d*x)^m*Cot[e + f*x])/f, x]
+ Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cot[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{(e+f x) \sinh ^3(c+d x)}{a+i a \sinh (c+d x)} \, dx &=i \int \frac{(e+f x) \sinh ^2(c+d x)}{a+i a \sinh (c+d x)} \, dx-\frac{i \int (e+f x) \sinh ^2(c+d x) \, dx}{a}\\ &=-\frac{i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f \sinh ^2(c+d x)}{4 a d^2}+\frac{i \int (e+f x) \, dx}{2 a}+\frac{\int (e+f x) \sinh (c+d x) \, dx}{a}-\int \frac{(e+f x) \sinh (c+d x)}{a+i a \sinh (c+d x)} \, dx\\ &=\frac{i e x}{2 a}+\frac{i f x^2}{4 a}+\frac{(e+f x) \cosh (c+d x)}{a d}-\frac{i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f \sinh ^2(c+d x)}{4 a d^2}-i \int \frac{e+f x}{a+i a \sinh (c+d x)} \, dx+\frac{i \int (e+f x) \, dx}{a}-\frac{f \int \cosh (c+d x) \, dx}{a d}\\ &=\frac{3 i e x}{2 a}+\frac{3 i f x^2}{4 a}+\frac{(e+f x) \cosh (c+d x)}{a d}-\frac{f \sinh (c+d x)}{a d^2}-\frac{i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f \sinh ^2(c+d x)}{4 a d^2}-\frac{i \int (e+f x) \csc ^2\left (\frac{1}{2} \left (i c+\frac{\pi }{2}\right )+\frac{i d x}{2}\right ) \, dx}{2 a}\\ &=\frac{3 i e x}{2 a}+\frac{3 i f x^2}{4 a}+\frac{(e+f x) \cosh (c+d x)}{a d}-\frac{f \sinh (c+d x)}{a d^2}-\frac{i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f \sinh ^2(c+d x)}{4 a d^2}-\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}+\frac{(i f) \int \coth \left (\frac{c}{2}-\frac{i \pi }{4}+\frac{d x}{2}\right ) \, dx}{a d}\\ &=\frac{3 i e x}{2 a}+\frac{3 i f x^2}{4 a}+\frac{(e+f x) \cosh (c+d x)}{a d}+\frac{2 i f \log \left (\cosh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )\right )}{a d^2}-\frac{f \sinh (c+d x)}{a d^2}-\frac{i (e+f x) \cosh (c+d x) \sinh (c+d x)}{2 a d}+\frac{i f \sinh ^2(c+d x)}{4 a d^2}-\frac{i (e+f x) \tanh \left (\frac{c}{2}+\frac{i \pi }{4}+\frac{d x}{2}\right )}{a d}\\ \end{align*}

Mathematica [A]  time = 1.80156, size = 325, normalized size = 1.86 \[ \frac{\left (\cosh \left (\frac{1}{2} (c+d x)\right )+i \sinh \left (\frac{1}{2} (c+d x)\right )\right ) \left (\cosh \left (\frac{1}{2} (c+d x)\right ) \left (2 \left (-3 c^2 f-d (e+f x) \sinh (2 (c+d x))+6 c d e+4 i f \sinh (c+d x)+8 i f \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+4 f \log (\cosh (c+d x))-4 i c f+6 d^2 e x+3 d^2 f x^2-4 i d f x\right )-8 i d (e+f x) \cosh (c+d x)+f \cosh (2 (c+d x))\right )+\sinh \left (\frac{1}{2} (c+d x)\right ) \left (8 d (e+f x) \cosh (c+d x)+i \left (f \cosh (2 (c+d x))+2 \left (-3 c^2 f-d (e+f x) \sinh (2 (c+d x))+6 c d e+4 i f \sinh (c+d x)+8 i f \tan ^{-1}\left (\tanh \left (\frac{1}{2} (c+d x)\right )\right )+4 f \log (\cosh (c+d x))-4 i c f+6 d^2 e x+3 d^2 f x^2+8 i d e+4 i d f x\right )\right )\right )\right )}{8 a d^2 (\sinh (c+d x)-i)} \]

Antiderivative was successfully verified.

[In]

Integrate[((e + f*x)*Sinh[c + d*x]^3)/(a + I*a*Sinh[c + d*x]),x]

[Out]

((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*(Cosh[(c + d*x)/2]*((-8*I)*d*(e + f*x)*Cosh[c + d*x] + f*Cosh[2*(c
+ d*x)] + 2*(6*c*d*e - (4*I)*c*f - 3*c^2*f + 6*d^2*e*x - (4*I)*d*f*x + 3*d^2*f*x^2 + (8*I)*f*ArcTan[Tanh[(c +
d*x)/2]] + 4*f*Log[Cosh[c + d*x]] + (4*I)*f*Sinh[c + d*x] - d*(e + f*x)*Sinh[2*(c + d*x)])) + Sinh[(c + d*x)/2
]*(8*d*(e + f*x)*Cosh[c + d*x] + I*(f*Cosh[2*(c + d*x)] + 2*((8*I)*d*e + 6*c*d*e - (4*I)*c*f - 3*c^2*f + 6*d^2
*e*x + (4*I)*d*f*x + 3*d^2*f*x^2 + (8*I)*f*ArcTan[Tanh[(c + d*x)/2]] + 4*f*Log[Cosh[c + d*x]] + (4*I)*f*Sinh[c
+ d*x] - d*(e + f*x)*Sinh[2*(c + d*x)])))))/(8*a*d^2*(-I + Sinh[c + d*x]))

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Maple [A]  time = 0.109, size = 197, normalized size = 1.1 \begin{align*}{\frac{{\frac{3\,i}{4}}f{x}^{2}}{a}}+{\frac{{\frac{3\,i}{2}}ex}{a}}-{\frac{{\frac{i}{16}} \left ( 2\,dfx+2\,de-f \right ){{\rm e}^{2\,dx+2\,c}}}{a{d}^{2}}}+{\frac{ \left ( dfx+de-f \right ){{\rm e}^{dx+c}}}{2\,a{d}^{2}}}+{\frac{ \left ( dfx+de+f \right ){{\rm e}^{-dx-c}}}{2\,a{d}^{2}}}+{\frac{{\frac{i}{16}} \left ( 2\,dfx+2\,de+f \right ){{\rm e}^{-2\,dx-2\,c}}}{a{d}^{2}}}-{\frac{2\,ifx}{da}}-{\frac{2\,ifc}{a{d}^{2}}}+2\,{\frac{fx+e}{da \left ({{\rm e}^{dx+c}}-i \right ) }}+{\frac{2\,if\ln \left ({{\rm e}^{dx+c}}-i \right ) }{a{d}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x)

[Out]

3/4*I*f*x^2/a+3/2*I*e*x/a-1/16*I*(2*d*f*x+2*d*e-f)/a/d^2*exp(2*d*x+2*c)+1/2*(d*f*x+d*e-f)/a/d^2*exp(d*x+c)+1/2
*(d*f*x+d*e+f)/a/d^2*exp(-d*x-c)+1/16*I*(2*d*f*x+2*d*e+f)/a/d^2*exp(-2*d*x-2*c)-2*I*f/a/d*x-2*I*f/a/d^2*c+2*(f
*x+e)/d/a/(exp(d*x+c)-I)+2*I*f/a/d^2*ln(exp(d*x+c)-I)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 2.55633, size = 586, normalized size = 3.35 \begin{align*} \frac{2 \, d f x + 2 \, d e +{\left (-2 i \, d f x - 2 i \, d e + i \, f\right )} e^{\left (5 \, d x + 5 \, c\right )} +{\left (6 \, d f x + 6 \, d e - 7 \, f\right )} e^{\left (4 \, d x + 4 \, c\right )} +{\left (12 i \, d^{2} f x^{2} - 8 i \, d e +{\left (24 i \, d^{2} e - 40 i \, d f\right )} x + 8 i \, f\right )} e^{\left (3 \, d x + 3 \, c\right )} + 4 \,{\left (3 \, d^{2} f x^{2} + 10 \, d e + 2 \,{\left (3 \, d^{2} e + d f\right )} x + 2 \, f\right )} e^{\left (2 \, d x + 2 \, c\right )} +{\left (-6 i \, d f x - 6 i \, d e - 7 i \, f\right )} e^{\left (d x + c\right )} - 32 \,{\left (-i \, f e^{\left (3 \, d x + 3 \, c\right )} - f e^{\left (2 \, d x + 2 \, c\right )}\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + f}{16 \, a d^{2} e^{\left (3 \, d x + 3 \, c\right )} - 16 i \, a d^{2} e^{\left (2 \, d x + 2 \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="fricas")

[Out]

(2*d*f*x + 2*d*e + (-2*I*d*f*x - 2*I*d*e + I*f)*e^(5*d*x + 5*c) + (6*d*f*x + 6*d*e - 7*f)*e^(4*d*x + 4*c) + (1
2*I*d^2*f*x^2 - 8*I*d*e + (24*I*d^2*e - 40*I*d*f)*x + 8*I*f)*e^(3*d*x + 3*c) + 4*(3*d^2*f*x^2 + 10*d*e + 2*(3*
d^2*e + d*f)*x + 2*f)*e^(2*d*x + 2*c) + (-6*I*d*f*x - 6*I*d*e - 7*I*f)*e^(d*x + c) - 32*(-I*f*e^(3*d*x + 3*c)
- f*e^(2*d*x + 2*c))*log(e^(d*x + c) - I) + f)/(16*a*d^2*e^(3*d*x + 3*c) - 16*I*a*d^2*e^(2*d*x + 2*c))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)**3/(a+I*a*sinh(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.43709, size = 479, normalized size = 2.74 \begin{align*} \frac{12 i \, d^{2} f x^{2} e^{\left (3 \, d x + 4 \, c\right )} + 12 \, d^{2} f x^{2} e^{\left (2 \, d x + 3 \, c\right )} - 2 i \, d f x e^{\left (5 \, d x + 6 \, c\right )} + 6 \, d f x e^{\left (4 \, d x + 5 \, c\right )} + 24 i \, d^{2} x e^{\left (3 \, d x + 4 \, c + 1\right )} - 40 i \, d f x e^{\left (3 \, d x + 4 \, c\right )} + 24 \, d^{2} x e^{\left (2 \, d x + 3 \, c + 1\right )} + 8 \, d f x e^{\left (2 \, d x + 3 \, c\right )} - 6 i \, d f x e^{\left (d x + 2 \, c\right )} + 2 \, d f x e^{c} + 32 i \, f e^{\left (3 \, d x + 4 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) + 32 \, f e^{\left (2 \, d x + 3 \, c\right )} \log \left (e^{\left (d x + c\right )} - i\right ) - 2 i \, d e^{\left (5 \, d x + 6 \, c + 1\right )} + i \, f e^{\left (5 \, d x + 6 \, c\right )} + 6 \, d e^{\left (4 \, d x + 5 \, c + 1\right )} - 7 \, f e^{\left (4 \, d x + 5 \, c\right )} - 8 i \, d e^{\left (3 \, d x + 4 \, c + 1\right )} + 8 i \, f e^{\left (3 \, d x + 4 \, c\right )} + 40 \, d e^{\left (2 \, d x + 3 \, c + 1\right )} + 8 \, f e^{\left (2 \, d x + 3 \, c\right )} - 6 i \, d e^{\left (d x + 2 \, c + 1\right )} - 7 i \, f e^{\left (d x + 2 \, c\right )} + 2 \, d e^{\left (c + 1\right )} + f e^{c}}{16 \, a d^{2} e^{\left (3 \, d x + 4 \, c\right )} - 16 i \, a d^{2} e^{\left (2 \, d x + 3 \, c\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*sinh(d*x+c)^3/(a+I*a*sinh(d*x+c)),x, algorithm="giac")

[Out]

(12*I*d^2*f*x^2*e^(3*d*x + 4*c) + 12*d^2*f*x^2*e^(2*d*x + 3*c) - 2*I*d*f*x*e^(5*d*x + 6*c) + 6*d*f*x*e^(4*d*x
+ 5*c) + 24*I*d^2*x*e^(3*d*x + 4*c + 1) - 40*I*d*f*x*e^(3*d*x + 4*c) + 24*d^2*x*e^(2*d*x + 3*c + 1) + 8*d*f*x*
e^(2*d*x + 3*c) - 6*I*d*f*x*e^(d*x + 2*c) + 2*d*f*x*e^c + 32*I*f*e^(3*d*x + 4*c)*log(e^(d*x + c) - I) + 32*f*e
^(2*d*x + 3*c)*log(e^(d*x + c) - I) - 2*I*d*e^(5*d*x + 6*c + 1) + I*f*e^(5*d*x + 6*c) + 6*d*e^(4*d*x + 5*c + 1
) - 7*f*e^(4*d*x + 5*c) - 8*I*d*e^(3*d*x + 4*c + 1) + 8*I*f*e^(3*d*x + 4*c) + 40*d*e^(2*d*x + 3*c + 1) + 8*f*e
^(2*d*x + 3*c) - 6*I*d*e^(d*x + 2*c + 1) - 7*I*f*e^(d*x + 2*c) + 2*d*e^(c + 1) + f*e^c)/(16*a*d^2*e^(3*d*x + 4
*c) - 16*I*a*d^2*e^(2*d*x + 3*c))